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Unformatted text preview: S E C T I O N 16.2 Double Integrals over More General Regions (ET Section 15.2) 855 Case 1: a 1. Then in the interval of integration y 1. Also since T , we may assume that T &gt; 0. Thus, e T y y e T 1 1 = e T Hence, Z a 1 e T y y dy Z a 1 e T dy = e T ( a 1 ) By the limit lim T e T ( a 1 ) = 0 and the Squeeze Theorem we conclude that, lim T Z a 1 e T y y dy = Case 2: &lt; a &lt; 1. Then, Z a 1 e T y y dy = Z 1 a e T y y dy and in the interval of integration a y 1, therefore e T y y e T a a (the function e T y y is decreasing). Hence, Z 1 a e T y y dy Z 1 a e T a a dy = ( 1 a ) a e T a By the limit lim T 1 a a e T a = 0 and the Squeeze Theorem we conclude also that lim T Z a 1 e T y y = lim T Z 1 a e T y y = We thus showed that for all a &gt; 0, lim T Z a 1 e T y y = 0. Combining with Eq. (5) obtained in part (c), we find that I ( a ) = ln a . 16.2 Double Integrals over More General Regions (ET Section 15.2) Preliminary Questions 1. Which of the following expressions do not make sense? (a) Z 1 Z y 1 f ( x , y ) dy dx (b) Z 1 Z x 1 f ( x , y ) dy dx (c) Z y 1 Z 1 f ( x , y ) dy dx (d) Z 1 Z 1 x f ( x , y ) dy dx SOLUTION (a) This integral is the same as Z 1 Z y 1 f ( x , y ) dy dx = Z 1 Z y 1 f ( x , y ) dy dx The inner integral is an integral with respect to y , over the interval [1 , y ]. This does not make sense. (b) This integral is the following iterated integral: Z 1 Z x 1 f ( x , y ) dy dx = Z 1 Z x 1 f ( x , y ) dy dx The inner integral is a function of x and it is integrated with respect to x over the interval 0 x 1. The result is a number. This integral makes sense. 856 C H A P T E R 16 MULTIPLE INTEGRATION (ET CHAPTER 15) (c) This integral is the following iterated integral: Z y 1 Z 1 f ( x , y ) dy dx The inner integral is a function of x and it is integrated with respect to x over the interval 1 x y . The result is a function of y . This expression makes sense but the value of the integral is a function of y rather than a fixed number. (d) This integral is the following iterated integral: Z 1 Z 1 x f ( x , y ) dy dx The inner integral is a function of x and it is integrated with respect to x . This makes sense. 2. Draw a domain in the plane that is neither vertically nor horizontally simple. SOLUTION The following region cannot be described in the form a x b , ( x ) y ( x ) nor in the form c y d , ( y ) x ( y ) , hence it is neither vertically nor horizontally simple....
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 Spring '08
 Rogawski
 Integrals, dy, vertically simple region, Theorem 2

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